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Is m + z > 0 (1) m  3z > 0 (2) 4z  m > 0 [#permalink]
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Is m + z > 0 (1) m  3z > 0 (2) 4z  m > 0 [#permalink]
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Re: Is m + z > 0 (1) m  3z > 0 (2) 4z  m > 0 [#permalink]
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smitakokne wrote: Is m+z > 0
1. m3z > 0 2. 4zm > 0
OA : C
Need help in underdstanding how we arrive at C. Bunuel, is m+z > 0 the same as m/z > 1 ? if so, then 1. would be m/z > 3, which is SUFF and 2. would be m/z < 4, which is INSUFF so I would have said A Please correct me if I'm wrong  my inequality skills are a bit rusty



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Re: Is m + z > 0 (1) m  3z > 0 (2) 4z  m > 0 [#permalink]
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psirus wrote: smitakokne wrote: Is m+z > 0
1. m3z > 0 2. 4zm > 0
OA : C
Need help in underdstanding how we arrive at C. Bunuel, is m+z > 0 the same as m/z > 1 ? if so, then 1. would be m/z > 3, which is SUFF and 2. would be m/z < 4, which is INSUFF so I would have said A Please correct me if I'm wrong  my inequality skills are a bit rusty No, it's not correct. When you are writing m/z>1 from m+z>0 you are actually dividing both parts of inequality by z: never multiply or reduce (divide) inequality by an unknown (a variable) unless you are sure of its sign since you do not know whether you must flip the sign of the inequality.So if we knew that z>0 then m+z>0 > m/z+1 >0 and if we knew that z<0 then m+z>0 > m/z+1 <0. Hope it's clear.
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Re: Is m + z > 0 (1) m  3z > 0 (2) 4z  m > 0 [#permalink]
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19 Jan 2011, 22:14
Bunuel wrote: smitakokne wrote: Is m+z > 0
1. m3z > 0 2. 4zm > 0
OA : C
Need help in underdstanding how we arrive at C. (1) m  3z > 0. Insufficient on its own. (2) 4z  m > 0. Insufficient on its own. (1)+(2) Remember we can add inequalities with the sign in the same direction > \(m3z+4zm>0\) > \(z>0\), so \(z\) is positive. From (1) \(m>3z=positive\), so \(m\) is positive too (\(m\) is more than some positive number \(3z\), so it's positive) > \(m+z=positive+positive>0\). Sufficient. Answer: C. Hello Bunnel, As you said in the above post.. "(1)+(2) Remember we can add inequalities with the sign in the same direction >"what should we have different signs...



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Re: Is m + z > 0 (1) m  3z > 0 (2) 4z  m > 0 [#permalink]
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20 Jan 2011, 02:19
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jullysabat wrote: Bunuel wrote: smitakokne wrote: Is m+z > 0
1. m3z > 0 2. 4zm > 0
OA : C
Need help in underdstanding how we arrive at C. (1) m  3z > 0. Insufficient on its own. (2) 4z  m > 0. Insufficient on its own. (1)+(2) Remember we can add inequalities with the sign in the same direction > \(m3z+4zm>0\) > \(z>0\), so \(z\) is positive. From (1) \(m>3z=positive\), so \(m\) is positive too (\(m\) is more than some positive number \(3z\), so it's positive) > \(m+z=positive+positive>0\). Sufficient. Answer: C. Hello Bunnel, As you said in the above post.. "(1)+(2) Remember we can add inequalities with the sign in the same direction >"what should we have different signs... You can only add inequalities when their signs are in the same direction:If \(a>b\) and \(c>d\) (signs in same direction: \(>\) and \(>\)) > \(a+c>b+d\). Example: \(3<4\) and \(2<5\) > \(3+2<4+5\). You can only apply subtraction when their signs are in the opposite directions:If \(a>b\) and \(c<d\) (signs in opposite direction: \(>\) and \(<\)) > \(ac>bd\) (take the sign of the inequality you subtract from). Example: \(3<4\) and \(5>1\) > \(35<41\).
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Re: Is m + z > 0 (1) m  3z > 0 (2) 4z  m > 0 [#permalink]
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26 May 2011, 10:10
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Plus in arbitary values to check is m+z is always greater than 0 or there are other possibilities
st1 ==> m  3z >0 m =1 and z = 1 ==> m+z = 0 m=2 and z = 1 ==> m+z >0 not sufficient
st2 ==> 4zm>0 m = 1 and z = 0.1 ==> m+z < 0 m=2 and z = 1 ==> m+z >0 not sufficient
combining both
m3z + 4z m >0 z > 0.... (P) from 1, m > 3z and from 2, 4z > m ==> 3z < m < 4z... (Q) from (P) and (Q) m+Z > 0
Answer C.
Last edited by agdimple333 on 26 May 2011, 11:13, edited 1 time in total.



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Re: Is m + z > 0 (1) m  3z > 0 (2) 4z  m > 0 [#permalink]
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Wow  Thank you agdimple333. That was quick and clear! I was combining both to get that 3z < m < 4z but I did not add them to get that z > 0 as well. Now that I think again if m lies between 3z and 4z, then by this equation alone z has to be GT 0 because there is NO z < 0 that will ever satisfy this equation. Thanks again.
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Re: Is m + z > 0 (1) m  3z > 0 (2) 4z  m > 0 [#permalink]
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11 Aug 2011, 04:27
Brunel, could you help me to figure out where I am wrong.
From statement 1: M3Z > 0
Adding 3Z to both sides of inequality
M>3Z
is Z is negative, then 3 times that negative number is definitely less than that number
for example Z = .25
3Z = .75 < Z
Z = 1 3Z =  3 < Z
Zero does not apply because, then both sides would be equal.
Since we its evident that both Z and M are positive Z+M is greater than zero...
Where am I going wrong with my reasoning?



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Re: Is m + z > 0 (1) m  3z > 0 (2) 4z  m > 0 [#permalink]
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11 Aug 2011, 07:27
manishgeorge wrote: Brunel, could you help me to figure out where I am wrong.
From statement 1: M3Z > 0
Adding 3Z to both sides of inequality
M>3Z
is Z is negative, then 3 times that negative number is definitely less than that number
for example Z = .25
3Z = .75 < Z
Z = 1 3Z =  3 < Z
Zero does not apply because, then both sides would be equal.
Since we its evident that both Z and M are positive Z+M is greater than zero...
Where am I going wrong with my reasoning? m>3z m=0; z=1; m>3z; m+z=1<0 m=1; z=0; m>3z; m+z=1>0 m=+1; z=1; m>3z; m+z=0 Thus, knowing that m>3z is not sufficient to find whether m+z>0 Not Sufficient.
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Re: Is m + z > 0 (1) m  3z > 0 (2) 4z  m > 0 [#permalink]
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arvindbhat1887 wrote: Is m+z>0?
(1) m > 3z (2) m < 4z.
The answer is given as (C) Both statements together. I don't understand how. Can someone please explain? I think the answer is (E) Not sufficient with both (1) tells you that m > 3z (2) tells you that m < 4z. Either of the two cases taken individually is not sufficient. So Rule out (A), (B), (D) Therefore, 3z<m<4z when you combine the two. Now z can take both ve and +ve values. So, m + z can be either ve or +ve depending on the value of z. Hence, (E). Please help. Stem 1: m>3z or m3z >0 doesnt tell us anything about m+z. Not sufficient. Stem 2: m<4z or m4z <0 or 4zm >0 tells us nothing again. Not sufficient. combining, we get that 3z <m <4z also, if we add both equations, m3z >0 4zm >0 we get, z>0 Thus since m>3z => m is also >0 therefore m+z >0 Sufficient. Ans C it is.
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Re: Is m + z > 0 (1) m  3z > 0 (2) 4z  m > 0 [#permalink]
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21 Jul 2013, 01:29
Bunuel wrote: (1) m  3z > 0. Insufficient on its own. (2) 4z  m > 0. Insufficient on its own. (1)+(2) Remember we can add inequalities with the sign in the same direction > \(m3z+4zm>0\) > \(z>0\), so \(z\) is positive. From (1) \(m>3z=positive\), so \(m\) is positive too (\(m\) is more than some positive number \(3z\), so it's positive) > \(m+z=positive+positive>0\). Sufficient. Answer: C. For graphic approach refer to: ismz01m3z024zm75657.htmlI have one question when statements are combined, I got till the part z > 0 now I got confused here since in statement 1 I could prove it m>3z ( m is positive) but didn't how to apply z>0 to the second statement >>> 4z  m > 0 So when you reach that stage you can apply it to any of the statement and conclude its sufficient or you have to conclude using both statement separately?
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Re: Is m + z > 0 (1) m  3z > 0 (2) 4z  m > 0 [#permalink]
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21 Jul 2013, 01:37
fozzzy wrote: Bunuel wrote: (1) m  3z > 0. Insufficient on its own. (2) 4z  m > 0. Insufficient on its own. (1)+(2) Remember we can add inequalities with the sign in the same direction > \(m3z+4zm>0\) > \(z>0\), so \(z\) is positive. From (1) \(m>3z=positive\), so \(m\) is positive too (\(m\) is more than some positive number \(3z\), so it's positive) > \(m+z=positive+positive>0\). Sufficient. Answer: C. For graphic approach refer to: ismz01m3z024zm75657.htmlI have one question when statements are combined, I got till the part z > 0 now I got confused here since in statement 1 I could prove it m>3z ( m is positive) but didn't how to apply z>0 to the second statement >>> 4z  m > 0 So when you reach that stage you can apply it to any of the statement and conclude its sufficient or you have to conclude using both statement separately? You got that m is positive with (1), so can stop there. If you combine you get that 4z>m>3z>0.
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Re: Is m + z > 0 (1) m  3z > 0 (2) 4z  m > 0 [#permalink]
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24 Nov 2013, 08:10
Bunuel wrote: smitakokne wrote: Is m+z > 0
1. m3z > 0 2. 4zm > 0
OA : C
Need help in underdstanding how we arrive at C. (1) m  3z > 0. Insufficient on its own. (2) 4z  m > 0. Insufficient on its own. (1)+(2) Remember we can add inequalities with the sign in the same direction > \(m3z+4zm>0\) > \(z>0\), so \(z\) is positive. From (1) \(m>3z=positive\), so \(m\) is positive too (\(m\) is more than some positive number \(3z\), so it's positive) > \(m+z=positive+positive>0\). Sufficient. Answer: C. For graphic approach refer to: ismz01m3z024zm75657.htmlHi Bunuel, When I took both statements and in my head made the stipulations, I ended up with 3z<m<4z, but that did not yield the right answer. Can you tell me what I am missing doing this vs. actually adding the equations?



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Re: Is m + z > 0 (1) m  3z > 0 (2) 4z  m > 0 [#permalink]
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24 Nov 2013, 08:13
ronr34 wrote: Bunuel wrote: smitakokne wrote: Is m+z > 0
1. m3z > 0 2. 4zm > 0
OA : C
Need help in underdstanding how we arrive at C. (1) m  3z > 0. Insufficient on its own. (2) 4z  m > 0. Insufficient on its own. (1)+(2) Remember we can add inequalities with the sign in the same direction > \(m3z+4zm>0\) > \(z>0\), so \(z\) is positive. From (1) \(m>3z=positive\), so \(m\) is positive too (\(m\) is more than some positive number \(3z\), so it's positive) > \(m+z=positive+positive>0\). Sufficient. Answer: C. For graphic approach refer to: ismz01m3z024zm75657.htmlHi Bunuel, When I took both statements and in my head made the stipulations, I ended up with 3z<m<4z, but that did not yield the right answer. Can you tell me what I am missing doing this vs. actually adding the equations? You missed the last step: from 3z<4z it follows that z>0, thus 3z=positive > (3z=positive)<m > m=positive > m+z=positive.
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Re: Is m + z > 0 (1) m  3z > 0 (2) 4z  m > 0 [#permalink]
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23 Jul 2014, 12:15
Is that correct?
1) \(m3z>0\) <=> \(m>3z\) Insuff 2) \(4zm>0\) <=> \(m<4z\) Insuff
1+2) we multiply 1) by 4, \(4m>12z\) we multiply 2) by 3, \(3m<12z\) <=> \(3m>12z\)
As we can add two inequalities \(4m3m>0\) so \(m>0.\)
from1), \(m>3z\) from 2), \(m<4z\) <=> \(m>4z\)
As we can add two inequalities \(0>z\) so \(z>0\)
so \(m+z>0\)



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Re: Is m + z > 0 (1) m  3z > 0 (2) 4z  m > 0 [#permalink]
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Re: Is m + z > 0 (1) m  3z > 0 (2) 4z  m > 0 [#permalink]
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07 Sep 2014, 07:41
Hi Bunuel, I have a doubt in this question. it is asked whether m+z>0 this implies that is m>z?
Statement 1 says: m3Z>0 then m>3z and 3z is definitely greater than z then isnt m>z. Please clarify.



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Re: Is m + z > 0 (1) m  3z > 0 (2) 4z  m > 0 [#permalink]
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Re: Is m + z > 0 (1) m  3z > 0 (2) 4z  m > 0 [#permalink]
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Bunuel, Skywalker18, Engr2012, and other experts : Do you think algebraic approach works even if an answer choice is E in such similar questions? Or do we have to resort to graphic approach/number picking then? Any tips on how to attack these questions in general? Thanks
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